Answer
$y'= cos^2(x^4)4x^3$
Work Step by Step
We are told to find the derivative using Part 1 of the Fundamental Theorem of Calculus given the expression: $y = \int ^{x^4}_{0}cos^2(θ)dθ$
Since the upper bound is a function we need to use the chain rule:
$u = x^4$
$u' = 4x^3$
$y = \int^u_0 cos^2(u)du$
Since $F(u) = \int^{u}_{0}f(t)dt$
$F'(u) = f(u)$
Thus,
$y' = cos^2(u)u'$
$y'= cos^2(x^4)4x^3$